## University Physics with Modern Physics (14th Edition)

Published by Pearson

# Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 59: 2.16

#### Answer

(a) average acceleration: magnitude: $1.0~m/s^2$ algebraic sign: negative direction: to the left (b) average acceleration: magnitude: $1.0~m/s^2$ algebraic sign: negative direction: to the left (c) average acceleration: magnitude: $3~m/s^2$ algebraic sign: negative direction: to the left

#### Work Step by Step

average acceleration = $\frac{\Delta v}{\Delta t}$ (a) $\frac{\Delta v}{\Delta t} = \frac{5.0~m/s - 15.0~m/s}{10~s} = -1.0~m/s^2$ average acceleration: magnitude: $1.0~m/s^2$ algebraic sign: negative direction: to the left (b) $\frac{\Delta v}{\Delta t} = \frac{-15.0~m/s - (-5.0)~m/s}{10~s} = -1.0~m/s^2$ average acceleration: magnitude: $1.0~m/s^2$ algebraic sign: negative direction: to the left (c) $\frac{\Delta v}{\Delta t} = \frac{-15.0~m/s - 15.0~m/s}{10~s} = -3.0~m/s^2$ average acceleration: magnitude: $3~m/s^2$ algebraic sign: negative direction: to the left

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